Question: Factor completely. $3x^2-147=$
Answer: First, we take a common factor of $3$. $3x^2-147=3(x^2-49)$ Now, let's factor $x^2-49$. Both $x^2$ and $49$ are perfect squares, since $x^2=({x})^2$ and $49=({7})^2$. $x^2-49 = ({x})^2-({7})^2$ So we can use the difference of squares pattern to factor. ${a}^2 - {b}^2 =({a}+{b})({a}-{b})$ In this case, ${a}={x}$ and ${b}={7}$ : $({x})^2 - ({7})^2 =({x}+{7})({x}-{7})$ $\begin{aligned} 3x^2-147&=3(x^2-49) \\\\ &=3(x+7)(x-7) \end{aligned}$ In conclusion, the complete factorization is: $3(x+7)(x-7)$ Remember that you can always check your factorization by expanding it.